1. ## [SOLVED] Sine Question

On a sunny day, a tower casts a shadow 35.2m long. At the same time, a 1.3m parking meter that is nearby casts a shadow 1.8m long. How high is the tower to the nearest tenth of a metre?

Triangle formed by parking meter:
$c^2 = a^2 + b^2$
$c^2 = 1.3^2 + 1.8^2$
$c^2 = \frac{493}{100}$
$c = \frac{\sqrt{493}}{10}$

$SinA = \frac{O}{H}$
$SinA = \frac{1.3}{\frac{\sqrt{493}}{10}}$
$Sin^{-1}\frac{1.3}{\frac{\sqrt{493}}{10}} = A$
$35.84^{\circ} = A$

Triangle formed by tower (Total triangle):
$\frac{a}{SinA} = \frac{b}{SinB}$
$\frac{a}{Sin35.84} = \frac{37}{Sin54.16}$
$a = Sin35.84(\frac{37}{Sin54.16})$
$a = 26.7m$

*The $37m$ value was obtained by adding $35.2m$ and $1.8m$.
*The $54.16^{\circ}$ value was obtained by subtracting $90deg$ and $35.84deg$ from $180deg$.

Therefore, the height of the tower is $26.7m$.

Textbook answer: $25.4m$.

Where did I make the mistake?

2. Hi RogueDemon,

Actually, you made this way harder than it is.

You have two similar right triangles.

Set up a proportion using the ratios of the heights and shadows.

$\frac{bldg \:\: ht.}{1.3}=\frac{35.2}{1.8}$

3. Thanks very much!

### on a sunny day, a tower cast 30m long shadow and iron rod

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